Prove Frattini's argument

  • #1
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Homework Statement:
The problem is two parts:

i) Let ##X## be a finite ##G##-set , and let ##H \le G## act transitively on ##X##. Then ##G = HG_x## for each ##x \in X##.

ii) Show that the Frattini argument follows from i).
Relevant Equations:
Relevant equations:

Frattini Argument: Let ##K## be a normal subgroup of a finite group ##G##. If ##P## is a Sylow ##p##-subgroup of ##K## (for some prime ##p##), then $$G = KN_G(P).$$

##X## is a finite ##G##-set means that ##G## acts on ##X## and both ##X## and ##G## are finite.

##G_x = \lbrace g \in G : g\cdot x = x \rbrace##
Attempt at solution:

Proof of i): Let ##x \in X##. Its clear ##G \supseteq HG_x##. Let ##g \in G##.Then there is ##y \in X## such that ##g \cdot x = y##. Since ##H## acts transitively on ##X##, there is ##h \in H## such that ##h \cdot x = y##. So, ##g \cdot x = h \cdot x##. This gives $$(h^{-1}g)\cdot x = x$$ Hence, ##g = h(h^{-1}g) \in HG_x## and we can conclude ##G = HG_x##. []

For ii), let ##X = \lbrace gPg^{-1} : g \in G\rbrace##. Then ##G## acts on ##X## by conjugation and ##N_G(P)## is the stabilizer of ##P##. But I'm not sure if ##K## acts transitively on ##X##. I know I haven't used the fact that ##K## is a normal subgroup of ##G##. Can I have a hint on how to solve ii), please?
 

Answers and Replies

  • #2
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I think I got it?

Proof of ii): Let ##X = \lbrace gPg^{-1} : g \in G \rbrace##. Then ##G## acts on ##X## by conjugation. If ##g \in G##, then ##gPg^{-1} \le gKg^{-1} = K## since ##K## is a normal subgroup of ##G##. Hence, ##gPg^{-1}## is a Sylow p-subgroup of ##K##. So there is ##k \in K## such that ##kPk^{-1} = gPg^{-1}##. In other words, ##K## acts transitively on ##X##. Moreover, ##N_G(P)## is the stabilizer of ##P##. By i),
$$G = KN_G(P).$$
[]
 
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